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Answer. Right ascension=198°.34'.32", and declination= 10.4.31" South.

3. Required the right ascension and declination of Betelguese, in the east shoulder of Orion; latitude 16°.3'.3" South, longitude 2.25°.51'.46"; obliquity of the ecliptic 23°.28′ ?

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Answer. Right ascension 85°.59'.28", declination=7°.21.17' North.

4. Required the right ascension and declination of Antares in the heart of Scorpio; latitude 4°.32.35" S., longitude 8.6°.51′.51"; obliquity of the ecliptic 23°.28'?

Answer. Right ascen. 244°.11.3", declination=25°.57′.22" South.

5. Required the right ascension and declination of the moon, when her latitude is 4°.0'.34" N., longitude 7.14°.26′.21′′, and the obliquity of the ecliptic 23°.27'.48"?

Answer. Right ascen. 223°.11.11", declin. -12°.21'.14" South.

6. Given the right ascension of Aldebaran=4.25'.12", and its declination 16°.7.27' N. in the year 1813, required its latitude and longitude; the obliquity of the ecliptic being 23°.28'? In the triangle nnc.

NNC=156°.18'

=AR+90°, AR being 4".25'.12".
NC = 73°.52′.33" complement of RC.
Nn = 23°28′. 0"=obliquity of the ecliptic.

1. To find the co-latitude nc.

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*The latitude of this star, on the same day, in the year 1796, was 5°.28′.27′′S. making a difference of only 25" in 17 years. (See the note page 271.)

2. To find the co-longitude cnn.

Sinenc 95°.28′.52′′:sinenNc=156°.18′:: sine NC=73°.52′.33": Sine cnN=22°.49'.28". Hence the longitude = 67°.10′.32′′ or 2$.7°.10'.32".

(L) By comparing together the places of the fixed stars deduced from observation, astronomers have found that their longitudes increase about 50" annually. This increase of longitude must necessarily cause an irregular motion in the same star with regard to the equinoctial. Hence the right ascensions, and declinations of stars are continually varying, so that stars which formerly had north declination have now south declination, et contra. Their latitudes are also subject to a small variation.

7. Required the latitude and longitude of Sirius in the Great Dog, right ascension being 99°.0.21", and declination= 16°.26'.35" South; the obliquity of the ecliptic being 23°.28'? Answer. Latitude 39°.337.30" S. longitude 3.11°.13'.44". 8. Required the latitude and longitude of Procyon in the Little Dog; right ascension being 1120.6.47", declination 5°.45'.3" N.; obliquity of the ecliptic 23°.28'?

Answer. Latitude 15°.58.14" S., longitude 3o.22°.55′,42′′. 9. Required the latitude and longitude of Arcturus in Boötes; right ascension 211°.33'.17", declination 20°.15′.58" N.; the obliquity of the ecliptic being 23°.28'?

Answer. Lat. 30°.52'.43" N. long. 6.21°.20.41".

10. Required the latitude and longitude of Fomalhaut, in the Southern Fish; right ascension being 341°.32′.34", declination 30°.42′.51′′ S.; obliquity of the ecliptic 23°.28'?

Answer. Lat. 21°.6′.29′′ S., long. 11.0°.56'.42".

11. Required the latitude and longitude of the moon when her right ascension is 40°.22′, declination 11°.37′ N., and the obliquity of the ecliptic 23°.28'?

Answer. Lat. 3°.53′ South, longitude 41°.35′.

Note. This problem may be varied so as to admit of several cases, but the two above specified are the only useful ones.

PROBLEM XI. (Plate 111. Fig. 8.)

(M) The right ascensions and declinations of two stars, or the latitudes and longitudes of two stars given, to find their distance.

Required the distance between Sirius in the Great Dog, right ascension=99°.0′.21′′, and declination 16°.26′.35′′ S.,

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The longitude of this star in the year 1796 was found, by the same process, so be 66°.56.24", making an increase of 14.8" in 17 years.

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and Procyon in the Little Dog, right ascension 112°.6'.47", and declination 5°.45'.3" N.?

In the triangle XST let the star at T represent Procyon, because its right ascension is the greatest, and the star at x represent Sirius.

The 4XST 13°.6'.26" diff. of their right ascensions=RQ. sx=73°.33′.25" the complement of Sirius's declination RX. ST=95. 45. 3 Procyon's declination +90°=sq+QT.

Hence, here are two sides and the contained angle given, to find the third side XT, by any of the methods in the preceding problems.

Answer. XT=95°.42′.10".

EXAMPLE II.

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The latitude and longitude of Sirius being 39°.33′.30′′ S., and 3.11°.13'.44"; and the latitude and longitude of Procyon 15.58.14" South, and 3.22°.55'.42". Required their distance?

In the triangle xmт as before, we have

xm 50°.26'.30" the complement of Sirius's lat. vx. Tm 74. 1.46 the complement of Procyon's lat. T. The 4xmT=11°.41'. 58" the difference of their longitudes

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Hence as before, here are two sides and the contained angle given, to find the third side, XT=25°.42' answer.

EXAMPLE III.

Required the distance between Capella in the Goat, right ascension=75°.21'.19", declination=45°.46′.15′′ North, and Procyon in the Little Dog, right ascension=112°.6′.47′′, and declination=5°.45'.3" North?

In the triangle TNC, let the star at T represent Procyon, and c Capella.

NC 44°.13.45" the co-declination of Capella.

TN 84. 14. 57. the co-declination of Procyon.

▲ TNC 36. 45. 28. the diff. of their right ascensions=RQ. Here we have two sides and the contained angle given, to find the third side, as in the preceding example.

Answer. CT 51°.6′.56′′.

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Required the distance between Capella, whose latitude is 22°.51.57" North, longitude 25.18°.57'.57", and Procyon, latitude 150.58.14" South, longitude 3.22°.55′.42" ?

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105°.58.14"

Let the star at T represent Procyon, that

comp. Capella's lat. oc.

90° + Procyon's lat.=nT.

Tnc 33°.57.45"-diff. of their longitudes=0. Here we have two sides and the contained angle given, to find the third side CT=51°.6'.39" Answer.

PROBLEM XII. (Plate III. Fig. 8.)

(N) The places of two stars being given, and their distances from a third star, to find the place of this third star.

EXAMPLE I.

Suppose the distance of a comet or new star c, was measured by a sextant to be 65°.47.42", from Sirius the dog star, lat. 39°.33'.30" South, longitude 35.11°.13′.44"; and 51°.6′.56′′, from Proycon the little Dog, lat. 15°.58'.14" South, longitude 3o.220.55.42′′; it is required to find the latitude and longitude of this comet or star.

Let x represent the place of Sirius, or one of the known stars, T that of Procyon the other known star, having the greatest longitude, and c the place of the comet or unknown

star.

First. In the triangle xnт, we have given xn=129°.33'.30" Sirius's distance from the pole of the

Tn105. 58. 14

ZxNT 11.41.58

ecliptic nearest the comet or star; Procyon's distance from the same pole; The diff. of longitude between the given stars.

To find XT the distance between the two known stars= 25°.42'.10", and the angle nxT = 26°.42′.39": To find the former of these quantities, there are two sides and the included angle given, and to find the latter the three sides and one angle are given.

Secondly. In the triangle XTC, we have given

XT=25°.42′10′′) To find the CXT at the star x. By
TC 51. 6.56 any of the methods (M.200.or G. 227,)

and xc=65. 47.42 you will find CXT=49°.14'.

This angle being greater than nxt shews the longitude of the nxt=22°.31′.21′′=

star at C to be less than at x. Hence cXT

nxc.

Thirdly, In the triangle xnc, we have given

xn=129°.33′.30") To find the side nc the comet or star's

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nc=67°.8.26" the star's co-latitude, hence its latitude oc= 22°.51'.34", and xnc=22°.16′.42", and since the star's longitude has been shewn to be greater than the comet's; from Sirius's longitude=101°.13′.44" take 22°.16′.42′′, the remainder 78°.57'.2" is the unknown star's longitude.

EXAMPLE II.

Suppose the distance of an unknown star at x was measured to be 65°.47'.42", from a known star (Capella) at c having 22°.51'.57" North latitude, and 25.18°.57.57" of longitude; and its distance from another known star at T (Procyon)=25°.42.10", this star having 15°.58'.14" South latitude, and 3.22°.55.42" of longitude; required the latitude and longitude of the unknown star?

First. In the triangle cmT, we shall have cm=mo+oc=112°.51'.57" Capella's distance from the pole of the ecliptic nearest the comet, or unknown star.

CMT= 33.57.45

Tm= 74. 1.46

To find TC= 51. 6.56

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The diff. long. between Capella

and Procyon.

Procyon's distance from the pole of the ecliptic.

The distance between the known

stars; and the angle mcr=43°.37′.41′′.

Secondly. In the triangle XCT, we have given

xc=65°.47.42′′

XT=25, 42. 10

TC 51. 6. 56

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To find the angle XCT at the star c.

By any of the methods (M. 200. or G. 227.), XCT will be found=24°.58'; this angle being less than mcr shews the longitude of the star at x to be greater than that at c.

Hence mcr

XCT-18°.39.41"-xcm.

Thirdly. In the triangle xcm, we have

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has

xm=50°.26′ the star's co-lat., hence its latitude vx=39°.34', and xmc=22°.15′.6", and since the given star's longitude been shewn to be less than the required star's longitude, this difference must be added to Capella's longitude, viz. to 2.18°.57.57" add 22°.15′.6" the sum is 3.11.13.3" the longitude required.

EXAMPLE

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